Let $n$ be a natural number and $a,b,c\in\Bbb R$.
How to calculate following determinant?
$$
\begin{vmatrix}
a & b & b & b & \dots & b & b \\
c & a & b & b & \dots & b & b \\
c & c & a & b & \dots & b & b \\
\vdots & \ddots&\ddots& \ddots &\ddots &\ddots& \vdots \\
c & c & c & c &\dots & a & b \\
c & c & c & c & \dots & c & a
\end{vmatrix}
$$
I. e. matrix has only $a$s in the diagonal, only $b$s in the top right and only $c$s in the bottom left.
I tried to develop the first row/column but te calculations don't lead anywhere. Please give me some help.
Best Answer
Subtract the 2nd column from the 1st, the 3rd from the 2nd, the 4th from the 3rd, &c. We obtain the determinant $$D_n=\begin{vmatrix} a-b&0&0&0&\dots &0&b\\ c-a& a-b&0&0&\dots &0&b \\ 0&c-a& a-b&0& \dots &0&b \\[-1ex] \vdots &&&\ddots&&\vdots\\ 0&0&0&0&\dots&a-b&b \\ 0&0&0&0&\dots&c-a&a \end{vmatrix}$$ Expand along the 1st row, noting the $(1,1)$ cofactor is just $D_{n-1}$ and the $(1,n)$ cofactor is upper triangular: $$D_n =(a-b)D_{n-1}+(-1)^{n-1}b(c-a)^{n-1}=(a-b)D_{n-1}+b(a-c)^{n-1}$$ On the other hand, swapping $b$ and $c$ changes the original matrix into its transpose. So in the above relation, we can swap $b$ and $c$, and $D_n$ also satisfies the relation $$D_n=(a-c)D_{n-1}+c(a-b)^{n-1},$$ whence by subtraction, $(b-c)D_{n-1}=b(a-c)^{n-1}-c(a-b)^{n-1}$, and ultimately $$D_{n-1}=\frac{b(a-c)^{n-1}-c(a-b)^{n-1}}{b-c}\qquad\text{ if }b\ne c$$
Case $\;b=c\,$:
Consider the numerator of $D_n$, for the case $b\ne c$ as a function of $x=c$: $$f(x)=b(a-x)^n-x(a-b)^n.$$ Note that $f(b)=0$, so that $D_n$ is $$D_n=-\frac{f(c)-f(b)}{c-b},\quad\text{ which tends to }\; -f'(b)\;\text{ when }\;c\to b.$$ So by continuity of the determinant,in the case $b=c$, we have \begin{align}D_n &=nb(a-x)^{n-1}+(a-b)^n\bigg|_{x=b}=nb(a-b)^{n-1}+(a-b)^n\\[1ex] &=(a-b)^{n-1}\bigl(a+(n-1)b\bigr). \end{align}